Question

A body is dropped from a height of 320m. The acceleration due to gravity is 10m/sec^2. (a) How long does it take to reach the ground. (b) What is the velocity with which it will strike the ground .

Answer 1

s=320m
a=10m/sec^2
u=0m/sec
t=?
v=?
we know s=ut+1/2at^2
320=(0)t+1/2(10)t^2
320×2=(10)t^2
640/10=t^2
√64=t
t=4sec
now v=?
we know
v=u+at
v=0+(10)4
v=40m/sec

Answer 2

"The velocity at which the body strikes the ground is $80\frac { m }{ sec }$ and time taken to reach the ground is 8 seconds.

Solution:

Consider a body is dropped from a height of 320m. The "acceleration due to gravity" is $10\frac { m }{ { sec }^{ 2 } }$ .

Given,

A body dropped from a height 's' = 320m

Acceleration 'a' $= 10\frac { m }{ { sec }^{ 2 } }$

initial velocity 'u' = $0\frac { m }{ sec }$    

Time 't' =?

Final velocity 'v'=?

we know $s\quad =\quad ut+\frac { 1 }{ 2 } a{ t }^{ 2 }$

By substituting the values in the above equation, we get.

$320\quad =\quad (0)t+\frac { 1 }{ 2 } (10){ t }^{ 2 }$

$320\times 2\quad =\quad (10){ t }^{ 2 }$

$\frac { 640 }{ 10 } \quad =\quad { t }^{ 2 }$

$\sqrt { 64 } \quad =\quad t$

t = 8sec

now v =?

we know

v = u + at

v = 0 + (10)8

$v = 80\frac { m }{ sec }$

The time taken to "reach the ground" is 8 second."